Question Number 191811 by mathlove last updated on 01/May/23
$$\int{x}^{\mathrm{2}} {e}^{−{x}} {dx}=? \\ $$
Answered by Spillover last updated on 01/May/23
$${use}\:{by}\:{parts} \\ $$
Answered by mr W last updated on 01/May/23
![∫x^2 e^(−x) dx =−∫x^2 d(e^(−x) ) =−x^2 e^(−x) +2∫xe^(−x) dx =−x^2 e^(−x) −2∫xd(e^(−x) ) =−x^2 e^(−x) −2[xe^(−x) −∫e^(−x) dx] =−x^2 e^(−x) −2xe^(−x) +2∫e^(−x) dx =−x^2 e^(−x) −2xe^(−x) −2e^(−x) +C =−(x^2 +2x+2)e^(−x) +C](https://www.tinkutara.com/question/Q191826.png)
$$\int{x}^{\mathrm{2}} {e}^{−{x}} {dx} \\ $$$$=−\int{x}^{\mathrm{2}} {d}\left({e}^{−{x}} \right) \\ $$$$=−{x}^{\mathrm{2}} {e}^{−{x}} +\mathrm{2}\int{xe}^{−{x}} {dx} \\ $$$$=−{x}^{\mathrm{2}} {e}^{−{x}} −\mathrm{2}\int{xd}\left({e}^{−{x}} \right) \\ $$$$=−{x}^{\mathrm{2}} {e}^{−{x}} −\mathrm{2}\left[{xe}^{−{x}} −\int{e}^{−{x}} {dx}\right] \\ $$$$=−{x}^{\mathrm{2}} {e}^{−{x}} −\mathrm{2}{xe}^{−{x}} +\mathrm{2}\int{e}^{−{x}} {dx} \\ $$$$=−{x}^{\mathrm{2}} {e}^{−{x}} −\mathrm{2}{xe}^{−{x}} −\mathrm{2}{e}^{−{x}} +{C} \\ $$$$=−\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right){e}^{−{x}} +{C} \\ $$
Commented by mathlove last updated on 01/May/23
$${thanks} \\ $$
Commented by mathlove last updated on 01/May/23
$${thanks} \\ $$
Commented by malwan last updated on 01/May/23
$${I}\:{think}\:{we}\:{can}\:{solve}\:{it}\:{by} \\ $$$${one}\:{step} \\ $$$${for}\:{example} \\ $$$$\int{x}^{\mathrm{3}} {e}^{−{x}} {dx}= \\ $$$$−{e}^{−{x}} \left({x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}\right)+{C} \\ $$